Consider the function {\f}:(0,2) \to {\R} defined by f(x)=\frac{x}{2}+\frac{2}{x} and the function g(x) defined by g(x)={\\begin{array}{c}\min \{f(t)\\end{array}, , 0< {\t} \leq {\x} \text{ and } 0< {\x} \leq 1 ; \frac{3}{2}+ {\x}, , 1< {\x}< 2}.. Then [27-Jan-2024 Shift 2]

Correct Answer is : g is continuous but not differentiable at x=1

g is continuous but not differentiable at x=1

g is not continuous for all x \in(0,2)

g is neither continuous nor differentiable at x=1

g is continuous and differentiable for all x \in(0,2)

JEE Main 27-Jan-2024 Shift 2 Solved Paper

Section: Mathematics

Question : 2 of 90

Marks: +1, -0

Consider the function f:(0,2)→R defined by f(x)=‌

+‌

and the function g(x) defined by g(x)={

min{f(t)}

‌

0<t≤x‌ and ‌0<x≤1

‌

1<x<2

.. Then

[27-Jan-2024 Shift 2]

Go to Question:

Prev Question

Next Question